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Prove sin(x) + cos(x) /=/ 1

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@Mihael_Keehl

cosx+sinx=2sin(x+π4)cosx+sinx=\sqrt{2}sin \left(x+\dfrac{\pi}{4} \right)

2sin(x+π4)=1\sqrt{2}sin \left(x+\dfrac{\pi}{4} \right)=1 only for certain values of x, certainly not for all xRx \in \mathbb{R}.
(edited 8 years ago)
Original post by Kvothe the arcane
@Mihael_Keehl

cosx+sinx=2sin(x+π4)cosx+sinx=\sqrt{2}sin \left(x+\dfrac{\pi}{4} \right)

2sin(x+π4)=1\sqrt{2}sin \left(x+\dfrac{\pi}{4} \right)=1 only for certain values of x, certainly not for all xRx \in \mathbb{R}.


nice thanks
Reply 42
Original post by Mihael_Keehl
...


Makes slightly more sense now, but it's a bit of a non question. It's like asking why x2+2x+10x^2 + 2x + 1 \neq 0. Sum of sinusoids are sinusoids themselves, so sinx+cosx\sin x + \cos x is a sinusoidal graph and hence varies, that is, it is not a constant (straight) line.
Original post by Zacken
Makes slightly more sense now, but it's a bit of a non question. It's like asking why x2+2x+10x^2 + 2x + 1 \neq 0. Sum of sinusoids are sinusoids themselves, so sinx+cosx\sin x + \cos x is a sinusoidal graph and hence varies, that is, it is not a constant (straight) line.


I see yea.

Why is it sinusoidal and not a combination of them both - and also is there such thing as a cosuiod?
Reply 44
Original post by Mihael_Keehl
I see yea.

Why is it sinusoidal and not a combination of them both - and also is there such thing as a cosuiod?


It is a combination of them both, but you need to realise that cosine is just sine shifted backwards by a bit, so it's still a sinusoid. Cosine is a translation of sine that comes in useful so often we gave it its own name.
Original post by Mihael_Keehl


I see.

So just by contradiction then?

Nothing we can do by induction or anytihng else.


*counterexample

If you can provide an example where it doesn't work, then you have proven the statement is false. I'm not sure whether there is a way of solving by induction or by another method, but there would be no point in doing that if you can just do proof by counterexample.
Original post by TimGB
*counterexample

If you can provide an example where it doesn't work, then you have proven the statement is false. I'm not sure whether there is a way of solving by induction or by another method, but there would be no point in doing that if you can just do proof by counterexample.


Yes I get that, thank you for your reply.

Original post by Zacken
It is a combination of them both, but you need to realise that cosine is just sine shifted backwards by a bit, so it's still a sinusoid. Cosine is a translation of sine that comes in useful so often we gave it its own name.

Thanks for the info.
Reply 47
Original post by Asklepios
True but that's not really a triangle is it. But I agree stupid question unless you exclude 0 and 2pi from the domain.


Non Euclidian geometry :biggrin:
Original post by Kvothe the arcane
32+42=523+4=53^2+4^2=5^2 \nRightarrow 3+4=5


Not applicable since these are trig functions but is thinking along the right lines.


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(edited 8 years ago)
Add to it's derivative and divide by 2. LOLOLOLOL
Then take the derivative again.
(edited 8 years ago)
Reply 50
This should work

Lets assume sinx + cos x = 1 and call it statement A

We also know sin^2 x + cos^x =1, which we'll call statement B

If we square statement A we should have (sin x + cos x)^2 = sin^2 x + cos^x +2sinx cos x = 1

minus statement B from statement A^2 leaves us wth 2sin x cos x = 0, which is only true when sin x or cos x = 0, which is also the time when sin x + cos x would equal 1.
I'm loving derivatives. How to prove sin^2(x) + cos^2(x) = 1? Just take the blimmin' derivative. (Circular reasoning for the circular functions)
Original post by physicsmaths
Not applicable since these are trig functions but is thinking along the right lines.


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I was just explaining what OP meant.

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